What is the integral of (sqrt(1/t-1)) 1/(t^2) ?

The integral of (sqrt(1/t-1)) 1/(t^2) is -2/3 ((1-t)/t)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (sqrt(1/t-1)) 1/(t^2)

\int (\frac{1}{t} - 1) \frac{1}{t ^{2}} d t

- \frac{2}{3} (\frac{1 - t}{t})^{\frac{3}{2}} + C

Solution steps

\int \frac{1}{t} - 1 \frac{1}{t ^{2}} d t

Apply u-substitution

= \int - u - 1 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u - 1 d u

Apply u-substitution

= - \int v d v

Apply the Power Rule

= - \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= - \frac{2}{3} (\frac{1}{t} - 1)^{\frac{3}{2}}

Join

\frac{1}{t} - 1 : \frac{1 - t}{t}

= - \frac{2}{3} (\frac{1 - t}{t})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (\frac{1 - t}{t})^{\frac{3}{2}} + C

What is the integral of (sqrt(1/t-1)) 1/(t^2) ?
The integral of (sqrt(1/t-1)) 1/(t^2) is -2/3 ((1-t)/t)^{3/2}+C